## Clocks Qns on relativity

1. The problem statement, all variables and given/known data

At what speed does a clock have to move in order to run at a rate which is one half that of a clock at rest?

2. Relevant equations

$$\Delta T = \gamma \Delta T_0$$

3. The attempt at a solution

For clocks to run at half it's spd,

$$\Delta T = 2 \Delta T_0$$

$$\gamma = 2$$

which works out to v = 0.866c

Is this correct?

Qns 2 :
i saw in the physic formulary list this :

Clock desynchronization: Moving clocks, synchronized in their rest frame but separated by a distance D along their direction of motion, are not synchronized in the stationary frame; The front clock lags the rear clock by an amount:
$$T = Dv/c^2$$

What does it mean?

If 2 clocks are moving at the same spd, shouldn't they run in the same time as well and should be synchronized?

Any help will be appreciated.
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 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus Your solution to the first question is correct. With respect to the second question. Imagine the clocks are synchronised in their rest frame, but are moving with respect to you. You are stood directly in between the two clocks (at D/2) such such that one is moving away from you at v m/s and one is approaching you at the same speed. Now, imagine that these clocks are digital and emit a flash every second. Which flash will you see first, the flash from the clock travelling away from you, or towards you?
 ohh......i get it, it has to do with the simultaneity of the events isn't it? So i will actually see the events of the front clock first before the rear clock. But one question, did the event actually happen at the same time? Is there a different simultaneity because light enter my eyes at different times although both events happen at the same time?

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## Clocks Qns on relativity

 Quote by Delzac ohh......i get it, it has to do with the simultaneity of the events isn't it?
Yes, its closely related.
 Quote by Delzac So i will actually see the events of the front clock first before the rear clock.
It's actually the other way round, but that's not the point.
 Quote by Delzac But one question, did the event actually happen at the same time? Is there a different simultaneity because light enter my eyes at different times although both events happen at the same time?
Both events are simultaneous in the clocks reference frame, but they are not simultaneous in your reference frame.
 But the isn't reason i don't see them as simultaneous because of the mystical property of light? Since light only helps us pass information only. Anyway, Clock2 ---------------me------------clock1 --------> direction of motion I believe that events that occur in clock1 will reach me first, is that correct?
 Part of the lecture slides :

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 Quote by Delzac But the isn't reason i don't see them as simultaneous because of the mystical property of light? Since light only helps us pass information only.
There is nothing 'mystical' about it, it is simply a Lorentz transformation.
 Quote by Delzac Clock2 ---------------me------------clock1 --------> direction of motion I believe that events that occur in clock1 will reach me first, is that correct?
No, that is not correct. Lets set up two reference frames. Let the S' frame be the rest frame of the clocks; let the S frame be your rest frame such that the S' frame is moving with a velocity v from left to right in your diagram. So, by the Lorentz transformation we have;

$$\Delta t = \gamma\left(\Delta t' +\frac{v\Delta x'}{c^2}\right)$$

Let us define $\Delta t:=t_2-t_1$, where t1 t2 are the times when clocks one and two emit their flashes respectively in the S frame. Similarity, $\Delta t':=t_2'-t_1'$, where t1' t2' are the times when clocks one and two emit their flashes respectively in the S' frame. Now, since the two clocks are synchronised in their rest frame (S'), the interval $\Delta t' = 0$. Thus we obtain;

$$\Delta t = \gamma\left(0 +\frac{v\Delta x'}{c^2}\right) = \Delta t = -\gamma\frac{v\Delta x'}{c^2}$$

Which is not only non-zero, but since $\Delta t < 0 \Rightarrow t_2-t_1 < 0 \Rightarrow t_2 < t_1$. Therefore, you will observe the flash from clock number 2 first.

Do you follow?
 Ok i get what you are saying, but that seem to contradict the picture from the lecture slide. what was in the picture was : Train moving in -------> direction lighting bolt struck-----------me--------------lighting bolt struck(front) And it wrote there that me(Mavis) saw the lighting bolt struck the front of the train first, then saw the LB struck the end of the train later. Is this true?
 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, the picture is correct, but is different from the clocks situation. In the train situation, mavis is moving towards event B and away from event A. However, in the clocks situation, clock 2 is moving towards you and clock 1 is moving away from you. Note, that in both cases you observe the event you are moving towards first. Does that make sense?
 Ahh.......yes it makes better sense now. But can i argue by saying that, well after all to see the event happening for the clock case, light must be transmitted to my eyes. Which is the same(maybe?) as the lighting bolt case since : direction of motion -------> Light from clock2---------> me <----------light from clock1
 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus No, it is not the same, it is the opposite as I have just explained. Consider what is moving towards you and what is moving away from you.
 Clock 1 is moving away from me Clock 2 is moving towards me. But can't it be that the light waves from clock 1 is moving also towards me?

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